Answer
(a) $1.4JK/s$
(b) less time
Work Step by Step
(a) We know that
$\frac{Q}{t}=(m_wc_w+m_sc_s)\frac{\Delta T}{t}$
We plug in the known values to obtain:
$\frac{Q}{t}=(\frac{(2.1L)(\frac{1Kg}{L})(4186)+(0.22)(448J/Kg.K)}{510s})(100C^{\circ}-22C^{\circ})$
$\implies \frac{Q}{t}=1.4KJ/s$
(b) The thermal conductivity of gold is $K_g=291W/m.K$ and the thermal conductivity of steel is $K_s=66.9W/m.K$. Therefore, the gold takes less time than the steel to boil the water.
